Ahelium-filled weather balloon has a volume of 793 l at 16.9°c and 759 mmhg. it is released and rises to an altitude of 4.05 km, where the pressure is 537 mmhg and the temperature is –7.1°c.

Assuming the question require we get the new volume at the altitude of 4.05 Km

1028.09 L

We are given;

Initial volume of Helium, V1 = 793 L

Initial temperature, T1 = 16.9 °C (But K = °C + 273.5)

, thus, T1 = 290.05 K                                

Initial pressure, P1 = 759 mmHg

New temperature at the height of 4.05 km, T2 =  -7.1°C + 273.15

                                    = 266.05 K

Pressure at the Height of 4.05 Km, P2 = 537 mmHg

We are required to calculate the volume attained by the balloon while at the height of 4.05 Km

We are going to use the combined gas law.

According to the combined gas law;

Therefore, making V2 the subject of the equation;

Therefore, the volume of the balloon at the height of 4.05 km will be 1028.09L

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.